pexp(120, 1/95, lower.tail = FALSE)
## [1] 0.2827597
PH = read.csv("PHISHING.csv")
hist(PH$INTTIME,main = "Phishing ")
hist(rexp(267, rate = 1/95))
The two graphs are very similar to each other.
Yes, the data appears to follow an exponential distribution.
alpha=3
beta=0.07
mean1=alpha*beta
mean1
## [1] 0.21
var1=alpha*beta^2
var1
## [1] 0.0147
alpha=3
beta=0.07
mean1=alpha*beta
var1=alpha*beta^2
lower=mean1-3*var1^0.5
upper=mean1+3*var1^0.5
paste("Interval is","(",lower,",",upper,")")
## [1] "Interval is ( -0.153730669589464 , 0.573730669589464 )"
alpha1=2
beta1=2
mean1=alpha1*beta1
var1=alpha1*beta1^2
mean1
## [1] 4
var1
## [1] 8
alpha2=1
beta2=4
mean2=alpha2*beta2
var2=alpha2*beta2^2
mean2
## [1] 4
var2
## [1] 16
#Formula A
p1=pgamma(1,shape=2,scale = 2)
p1
## [1] 0.09020401
#Formula B
p2=pgamma(1,shape=1,scale = 4)
p2
## [1] 0.2211992
Therefore, Formula B has higher probability of generating a human reaction in less than 1 minute.
pweibull(2,shape = 4, scale = 2)
## [1] 0.6321206
alpha=2
beta=4
mean1=beta^(1/alpha)*gamma((alpha+1)/alpha)
var1=beta^(2/alpha)*(gamma((alpha+2)/alpha)- (gamma((alpha+1)/alpha))^2)
sd1=sqrt(var1)
mean1
## [1] 1.772454
sd1
## [1] 0.9265028
alpha=2
beta=4
mean1=beta^(1/alpha)*gamma((alpha+1)/alpha)
var1=beta^(2/alpha)*(gamma((alpha+2)/alpha)- (gamma((alpha+1)/alpha))^2)
sd1=sqrt(var1)
lower=mean1-2*sd1
upper=mean1+2*sd1
paste("Interval is","(",lower,",",upper,")")
## [1] "Interval is ( -0.0805516497989005 , 3.62545935160993 )"
1-pweibull(6,shape = 4, scale = 2)
## [1] 0
No.
alpha=2
beta=9
mean1=alpha/(alpha+beta)
var1=(beta*alpha)/((alpha+beta)^2*(alpha+beta+1))
mean1
## [1] 0.1818182
var1
## [1] 0.01239669
Mean = 0.18182 Variance = 0.01240
1-pbeta(0.4,shape1 = 2, shape2 = 9)
## [1] 0.0463574
pbeta(0.1,shape1 = 2, shape2 = 9)
## [1] 0.2639011
alpha = 2
beta = 16
alpha=2
beta=16
mean=beta^(1/alpha)*gamma((alpha+1)/alpha)
var=beta^(2/alpha)*(gamma((alpha+2)/alpha)-(gamma((alpha+1)/alpha))^2)
mean
## [1] 3.544908
var
## [1] 3.433629
1-pweibull(6,shape=2,scale=16^(1/2))
## [1] 0.1053992
P(x,y)=\(\frac{1}{36}\)
\[ P_1(x) = \frac{1}{6}\\ P_2(y) = \frac{1}{6} \]
\[ \begin{align} P(x|y) &= \frac{p(x,y)}{p(y)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6} \end{align} \]
\[ \begin{align} P(y|x) &= \frac{p(x,y)}{p(x)}=\frac{\frac{1}{36}}{\frac{1}{6}}=\frac{1}{6} \end{align} \]
Since the conditional probabilities and the unconditional probabilities are equivalent, x and y don't affect each other.
8a
8b
\[ f(x,y)=\frac{e^{-y/10}}{10y},\;0<y<x<2y \]
For marginal density \(f(y)\)
\[ \begin{align} f(y)&=\int^\infty_{-\infty}f(x,y)dx \\ &= \int^{2y}_y\left[\frac{e^{-y/10}}{10y}\right]dx \\ &=\left[\frac{1}{10}\frac{e^{-y/10}}{y}\biggr\rvert^{2y}_y\right] \\ &= \frac{1}{10}e^{-y/10} \end{align} \]
to Find E(y) \[ \begin{align} E(y)&=\int_y y.f(y)dy\\ &=\int^\infty_0 \left[\frac{1}{10}e^{\frac{-y}{10}}\right]ydy\\ &=\left[\left[-(10+y)\right]\right]^\infty_0 \\ &= 10 \end{align} \]
10a